Temperature shows in cross section a wall consisting of four layers with thermal conductivities `K_(1)=0.06W//mK,K_(3)=0.04W//mK` and `K_(4)=0.10W//mK` the layer thickness are `L_(1)=1.5cm,L_(3)=2.8cm` and `L_(4)=3.5` cm. The temperature of interfaces is as shown in figure energy transfer through the wall is in steady state.
Q. Temperature of outer surface of sphere is

Options

A. `100^(@)`
B. `80^(@)`
C. `48^(@)`
D. `60^(@)`

Explanation

For spherical surface at steady state
`P_("heater")+50W=120W`
`impliesP_("Heater")=70W`
At steady state `P_("Heater")=((100-t_(out))kA)/(l)=70=k'(t_(out)-20)` ….(i)
From te given observation
`50=k'(40-20)`
from equation (ii) `k'=(5)/(2)`
From equation (i)
`t_(out)=48^(@)`
`l=0.52mm`